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CBSE NCERT Class 9

Chapter 2: Polynomials

An in-depth conceptual breakdown of polynomials in one variable, zeroes of a polynomial, Factor Theorem, and all 8 essential algebraic identities with applications.

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1. Definition & Degree of a Polynomial

An algebraic expression of the form:

General Form of Polynomial p(x)

p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀

Where aₙ, aₙ₋₁, ..., a₀ are real constants, and n is a whole number (non-negative integer).

Crucial Rule: The exponent of the variable must always be a non-negative integer (whole number: 0, 1, 2, 3...). Expressions like x^(1/2) + 5 or 1/x + 3 are NOT polynomials!

Degree of a Polynomial: The highest exponent of the variable x in a non-zero polynomial is called its degree.

Degree 0: Constant polynomial (e.g., p(x) = 7). Note: The degree of the zero polynomial 0 is not defined.
Degree 1: Linear polynomial (e.g., 2x + 3).
Degree 2: Quadratic polynomial (e.g., 4x² - 5x + 1).
Degree 3: Cubic polynomial (e.g., x³ - 2x² + x - 4).

2. Zeroes of a Polynomial

A real number c is called a zero of a polynomial p(x) if:

p(c) = 0

Key Theorems to remember:

A non-zero constant polynomial has no zeroes.
Every linear polynomial in one variable has a unique (one) zero.
A polynomial of degree n can have at most n real zeroes.

3. The Factor Theorem

If p(x) is a polynomial of degree n ≥ 1 and 'a' is any real number:

Factor Theorem Statement

1. If p(a) = 0, then (x - a) is a factor of p(x).
2. Conversely, if (x - a) is a factor of p(x), then p(a) = 0.

4. The 8 Standard Algebraic Identities

These 8 identities are fundamental to high school mathematics. Master each equation along with its factorised form:

Identity I

(x + y)² = x² + 2xy + y²

Identity II

(x - y)² = x² - 2xy + y²

Identity III

x² - y² = (x + y)(x - y)

Identity IV

(x + a)(x + b) = x² + (a + b)x + ab

Identity V (Trinomial)

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Identity VI (Cube Sum)

(x + y)³ = x³ + y³ + 3xy(x + y)

Identity VII (Cube Diff)

(x - y)³ = x³ - y³ - 3xy(x - y)

Identity VIII (Three Cubes)

x³ + y³ + z³ - 3xyz = (x+y+z)(x²+y²+z² - xy - yz - zx)

Special Exam Case for Identity VIII

If x + y + z = 0, then the identity simplifies directly to:
x³ + y³ + z³ = 3xyz

5. Step-by-Step Solved NCERT Examples

Exemplar Problem 1 Splitting the Middle Term

Question: Factorise 6x² + 17x + 5 by splitting the middle term.

Find product a × c and sum b:
Here a = 6, b = 17, c = 5. Product a × c = 6 × 5 = 30.
We need two numbers p and q such that p × q = 30 and p + q = 17.

Select factors: 15 × 2 = 30 and 15 + 2 = 17.
Split 17x as (15x + 2x):
= 6x² + 15x + 2x + 5

Group and take common factors:
= 3x(2x + 5) + 1(2x + 5)
= (2x + 5)(3x + 1).

Exemplar Problem 2 Evaluating Cubes Without Direct Calculation

Question: Without actually calculating the cubes, find the value of:
(-12)³ + (7)³ + (5)³

Let x = -12, y = 7, z = 5.
Check their sum: x + y + z = -12 + 7 + 5 = 0.

Since x + y + z = 0, we use Identity VIII: x³ + y³ + z³ = 3xyz.
= 3 × (-12) × 7 × 5
= -36 × 35 = -1260.

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